MCQ
$\int_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\right) \mathrm{dx}$=......................
- A$-1 / 2$
- B$1 / 4$
- C$1 / 2$
- D$-1 / 4$
$ \int_{1 / 4}^{3 / 4} \cos \left(2\left(\tan ^{-1} \sqrt{\frac{1+\mathrm{x}}{1+\mathrm{x}}}\right)\right) \mathrm{dx} $
$ \left.\int_{1 / 4}^{3 / 4} \frac{1-\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\right.}{1+\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\right.}\right) d x $
$ =\int_{1 / 4}^{3 / 4} \frac{1-\left(\frac{1+x}{1-x}\right)}{1+\left(\frac{1+x}{1-x}\right)} d x=\int_{1 / 4}^{3 / 4} \frac{-2 x}{2} d x $
$ =\int_{1 / 4}^{3 / 4}(-x) d x=-\left(\frac{x^2}{2}\right)_{1 / 4}^{3 / 4} $
$ =-\frac{1}{2}\left[\frac{9}{16}-\frac{1}{16}\right] $
$ =-\frac{1}{4} $
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